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SilverGirl
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Gauss' Law - Sphere with spherical cavity - Please help
A sphere of radius 2a is made of nonconducting material that has a uniform volume charge density . (Assume that the material does not affect the electric field.) A spherical cavity of radius a is now removed from the sphere, as shown in Figure P19.62. Show that the electric field within the cavity is uniform and is given by Ex = 0 and Ey = Pa/3Eo. (Hint: The field within the cavity is the superposition of the field due to the original uncut sphere, plus the field due to a sphere the size of the cavity with a uniform negative charge density -.)
http://greenlanternbattery.googlepages.com/p19-62.gif
P= Q/V
Esphere=Q/Eo
I'm not really sure if the Esphere is right, and I'm not sure where to go next if it is. In my notes from the prof, it says "for each point P in the cavity need to consider the contribution to the E field from the positive charge distribution (let the vector be r+) and from the negative charge distribution (r-)." I am not really sure what this means though.
Homework Statement
A sphere of radius 2a is made of nonconducting material that has a uniform volume charge density . (Assume that the material does not affect the electric field.) A spherical cavity of radius a is now removed from the sphere, as shown in Figure P19.62. Show that the electric field within the cavity is uniform and is given by Ex = 0 and Ey = Pa/3Eo. (Hint: The field within the cavity is the superposition of the field due to the original uncut sphere, plus the field due to a sphere the size of the cavity with a uniform negative charge density -.)
http://greenlanternbattery.googlepages.com/p19-62.gif
Homework Equations
P= Q/V
The Attempt at a Solution
Esphere=Q/Eo
I'm not really sure if the Esphere is right, and I'm not sure where to go next if it is. In my notes from the prof, it says "for each point P in the cavity need to consider the contribution to the E field from the positive charge distribution (let the vector be r+) and from the negative charge distribution (r-)." I am not really sure what this means though.
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